Problem: Solve for $X$. $X-\left[\begin{array}{rr}-20 & 10 \\ 3 & 14 \end{array}\right]=\left[\begin{array}{rr}13 & 18 \\ 1 & 19\end{array}\right] $ $X=$
The Strategy First, we can represent the matrices of the equation with letters, which will make the equation easier to handle. Then we can solve the equation for $X$ and obtain an expression with the letters we defined. Finally, we can substitute back the actual matrices into the resulting expression and simplify it. Solving the equation for $X$ We are given the following equation. $X-\left[\begin{array}{rr}-20 & 10 \\ 3 & 14 \end{array}\right]=\left[\begin{array}{rr}13 & 18 \\ 1 & 19\end{array}\right]$ Let's represent the above matrices as follows. $A=\left[\begin{array}{rr}-20 & 10 \\ 3 & 14 \end{array}\right] ~~~~~~~~~ B = \left[\begin{array}{rr}13 & 18 \\ 1 & 19\end{array}\right]$ Then we can rewrite the equation as follows. $X-A=B$ Now it's simple to solve the equation for $X$. $\begin{aligned}X-A&=B\\\\ X&=A+B \end{aligned}$ Finding $X$ We found that $X=A+B$. Now we can substitute the actual matrices back into the expression and simplify. $\begin{aligned}X&=A+B \\\\&=\left[\begin{array}{rr}-20 & 10 \\ 3 & 14 \end{array}\right]+\left[\begin{array}{rr}13 & 18 \\ 1 & 19\end{array}\right] \\\\\\&=\left[\begin{array}{rr}(-20+13) & (10+18) \\ (3+1) & (14+19) \end{array}\right] \\\\\\&=\left[\begin{array}{rr}-7 & 28 \\ 4 & 33\end{array}\right]\end{aligned}$ Summary $X=\left[\begin{array}{rr}-7 & 28 \\ 4 & 33\end{array}\right]$